GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Oct 2019, 02:36

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

S97-11

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58400
S97-11  [#permalink]

Show Tags

New post 16 Sep 2014, 01:51
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

38% (01:38) correct 62% (01:02) wrong based on 60 sessions

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58400
Re S97-11  [#permalink]

Show Tags

New post 16 Sep 2014, 01:51
1
Official Solution:

Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?

A. Zero
B. One
C. Two
D. Three
E. More than three


To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.

Three values that would make the set symmetrical are 0, 10, and 20:

{0, 4, 8, 12, 16}

{4, 8, 10, 12, 16}

{4, 8, 12, 16, 20}

We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).

Well, we can set up three scenarios, each with a relevant equation.

(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median:
\(\frac{40 + x}{5} = 8\)
\(40 + x = 40\)
\(x = 0\)

(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = x\)
\(40 + x = 5x\)
\(40 = 4x\)
\(x = 10\)

(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = 12\)
\(40 + x = 60\)
\(x = 20\)

We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.


Answer: D
_________________
Intern
Intern
avatar
B
Joined: 09 Aug 2016
Posts: 15
Re: S97-11  [#permalink]

Show Tags

New post 28 Aug 2016, 03:50
Bunuel wrote:
Official Solution:

Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?

A. Zero
B. One
C. Two
D. Three
E. More than three


To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.

Three values that would make the set symmetrical are 0, 10, and 20:

{0, 4, 8, 12, 16}

{4, 8, 10, 12, 16}

{4, 8, 12, 16, 20}

We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).

Well, we can set up three scenarios, each with a relevant equation.

(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median:
\(\frac{40 + x}{5} = 8\)
\(40 + x = 40\)
\(x = 0\)

(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = x\)
\(40 + x = 5x\)
\(40 = 4x\)
\(x = 10\)

(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = 12\)
\(40 + x = 60\)
\(x = 20\)

We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.


Answer: D







Hi,


Isnt it necessary for the values to be in AP,for the mean and median to be equal?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58400
Re: S97-11  [#permalink]

Show Tags

New post 28 Aug 2016, 05:41
ashima09 wrote:
Bunuel wrote:
Official Solution:

Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?

A. Zero
B. One
C. Two
D. Three
E. More than three


To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.

Three values that would make the set symmetrical are 0, 10, and 20:

{0, 4, 8, 12, 16}

{4, 8, 10, 12, 16}

{4, 8, 12, 16, 20}

We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).

Well, we can set up three scenarios, each with a relevant equation.

(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median:
\(\frac{40 + x}{5} = 8\)
\(40 + x = 40\)
\(x = 0\)

(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = x\)
\(40 + x = 5x\)
\(40 = 4x\)
\(x = 10\)

(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = 12\)
\(40 + x = 60\)
\(x = 20\)

We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.


Answer: D







Hi,


Isnt it necessary for the values to be in AP,for the mean and median to be equal?

___________________
No, consider {1, 1, 2, 2}.
_________________
Intern
Intern
avatar
Joined: 10 Apr 2018
Posts: 2
Re: S97-11  [#permalink]

Show Tags

New post 19 Apr 2018, 13:34
1
Hi,
It does not say anywhere, that the value of x should be positive or equal to zero.
This bring a confusion and reflects to the answer.


Kind regards
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58400
Re: S97-11  [#permalink]

Show Tags

New post 20 Apr 2018, 02:17
Intern
Intern
avatar
B
Joined: 23 Jun 2018
Posts: 3
Re: S97-11  [#permalink]

Show Tags

New post 24 Jun 2018, 17:55
Bunuel wrote:
Official Solution:

Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?

A. Zero
B. One
C. Two
D. Three
E. More than three


To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.

Three values that would make the set symmetrical are 0, 10, and 20:

{0, 4, 8, 12, 16}

{4, 8, 10, 12, 16}

{4, 8, 12, 16, 20}

We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).

Well, we can set up three scenarios, each with a relevant equation.

(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median:
\(\frac{40 + x}{5} = 8\)
\(40 + x = 40\)
\(x = 0\)

(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = x\)
\(40 + x = 5x\)
\(40 = 4x\)
\(x = 10\)

(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = 12\)
\(40 + x = 60\)
\(x = 20\)

We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.


Answer: D


Hello,
It says that the numbers need not be ascending order. So, why aren't we taking the case {4, 8, 16, 12, x} ?
which implies x= 40. Therefore, values of x is more than three.
Thanks in advance!
Intern
Intern
avatar
B
Joined: 20 May 2018
Posts: 16
Re: S97-11  [#permalink]

Show Tags

New post 12 Oct 2018, 09:09
ladhaa wrote:
Bunuel wrote:
Official Solution:

Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?

A. Zero
B. One
C. Two
D. Three
E. More than three


To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.

Three values that would make the set symmetrical are 0, 10, and 20:

{0, 4, 8, 12, 16}

{4, 8, 10, 12, 16}

{4, 8, 12, 16, 20}

We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).

Well, we can set up three scenarios, each with a relevant equation.

(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median:
\(\frac{40 + x}{5} = 8\)
\(40 + x = 40\)
\(x = 0\)

(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = x\)
\(40 + x = 5x\)
\(40 = 4x\)
\(x = 10\)

(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median:
\(\frac{40 + x}{5} = 12\)
\(40 + x = 60\)
\(x = 20\)

We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.


Answer: D


Hello,
It says that the numbers need not be ascending order. So, why aren't we taking the case {4, 8, 16, 12, x} ?
which implies x= 40. Therefore, values of x is more than three.
Thanks in advance!


In order to compute the median, values have to be in order (either ascending or descending). Therefore your statement is fallacious.
GMAT Club Bot
Re: S97-11   [#permalink] 12 Oct 2018, 09:09
Display posts from previous: Sort by

S97-11

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel






Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne